 Kristen Stokes

2022-07-08

Integrable function on [0,2] and its antiderivative
I got this question:
Let f be the integrable function defined on the interval [0,2] by the rule:

Does there exist real numbers c1 and c2 such that the function F defined below is an antiderivative for f (i.e. ${F}^{\prime }=f$)?

I got stuck since the function f got step/jump discontinuity at $x=1$. And I don't know how to proceed from there. Jenna Farmer

Expert

Step 1
Let's start with the original question as stated, and consider the requirement that ${F}^{\prime }=f$ everywhere on [0,2]. As we approach 1, the left and right derivatives (left and right limits in the definition of the derivative) do not approach the same limit, so while f(1) is defined, F′(1) is not, so in this instance, one cannot choose ${c}_{1}$, ${c}_{2}$ to satisfy the hypotheses.
Step 2
If instead of requiring ${F}^{\prime }=f$ everywhere on [0,2], we allow ${F}^{\prime }=f$ almost everywhere on [0,2] (i.e., on $\left[0,1\right)\cup \left(1,2\right]$), then we can note that since f is absolutely integrable, $F\left(x\right)={\int }_{0}^{x}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}\text{d}t$ is a continuous function of x on [0,2].
So to satisfy ${F}^{\prime }=f$, for this definition of F(x), on $\left[0,1\right)\cup \left(1,2\right]$, we could choose ${c}_{1}=0$, ${c}_{2}=-4/3$; more generally, we could choose ${c}_{1}\in \mathbb{R}$, ${c}_{2}={c}_{1}-\frac{4}{3}$. (These values would make the function F(x) as defined immediately above continuous.)
However, if we simply require ${F}^{\prime }=f$ on $\left[0,1\right)\cup \left(1,2\right]$ and do not ask F to be continuous, then as the answer above states, we could choose any real numbers ${c}_{1}$, ${c}_{2}$, which would lead to (at worst) F having a jump discontinuity at $x=1$ lilmoore11p8

Expert

Step 1
If when you say ${F}^{\prime }=f$ you mean ${F}^{\prime }\left(x\right)=f\left(x\right)$ for every $x\in \left[0,2\right]$ so that F is differentiable then note that F is not differentiable at $x=1$. So then $\mathrm{\forall }x\in \left[0,2\right]\setminus \left\{1\right\}$ we have ${F}^{\prime }\left(x\right)=f\left(x\right)$ so in this sense the equality is found for any ${c}_{1},{c}_{2}\in \mathbb{R}$.
Step 2
However there is a technical aspect that says these functions are not completely identical. Note that F′ is not defined at $x=1$ but f is defined at $x=1$, and no ${c}_{1},{c}_{2}$ can satisfy these needs.

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